A right-angled prism of refractive index μ _ 1 is placed in a rectangular block of refractive index μ _ 2 , which is surrounded by a medium of refractive index μ _ 3 , as shown in the figure. A ray of light e enters the

Correct answer: C. a-q;b-r;c-s;d-p; For e → i , 4 5 ^ ∘ > θ _ c 4 5 ^ ∘ > θ _ c 1 2 > μ _ 2 μ _ 1 μ _ 1 > 2 μ _ 2 For e → f , an angle of refraction is lesser than the angle of incidence. So, μ _ 2 > μ _ 1 and then μ _ 2 > μ _ 3 . For e → g, μ _ 1 = μ _ 2 . For e → h, μ _ 2 μ _ 3 .

JEE Advanced 2013Paper 2

Exam: JEE Advanced
Year: 2013
Paper: Paper 2
Subject: Physics
Chapter: Ray Optics

JEE Advanced 2013 · Paper 2 · Physics · Ray Optics

A right-angled prism of refractive index μ _ 1 is placed in a...

A right-angled prism of refractive index

{\mu}_{1}

is placed in a rectangular block of refractive index

{\mu}_{2}

, which is surrounded by a medium of refractive index

{\mu}_{3}

, as shown in the figure. A ray of light

e

enters the rectangular block at normal incidence. Depending upon the relationships between

{\mu}_{1}\text{,}{\mu}_{2}

and

{\mu}_{3}

, it takes one of the four possible paths

ef, eg, eh

ei

.
$A right-angled prism of refractive index μ _ 1 is placed in a rectangular block of refractive index μ _ 2 , which is$
Match the paths in List

I

with conditions of refractive indices in List

II

and select the correct answer using the codes given below the lists:

	List I		List II
A.	$e \to f$	P.	${\mu}_{1} > \sqrt{2} {\mu}_{2}$
B.	$e \to g$	Q.	${\mu}_{2} > {\mu}_{1} \text{and} {\mu}_{2} > {\mu}_{3}$
C.	$e \to h$	R.	${\mu}_{1} = {\mu}_{2}$
D.	$e \to i$	S.	${\mu}_{2} {\mu}_{3}$

Options

Aa-s;b-r;c-p;d-q;
Ba-q;b-s;c-r;d-p;
Ca-q;b-r;c-s;d-p;
Da-s;b-r;c-p;d-q;

Correct answer

C. a-q;b-r;c-s;d-p;

Step-by-step solution

For

e \to i

4 {5}^{∘} > {\theta}_{\text{c}}

\text\sin 4 {5}^{∘} > \text\sin {\theta}_{\text{c}}

\dfrac{1}{\sqrt{2}} > \dfrac{{\mu}_{2}}{{\mu}_{1}}

{\mu}_{1} > \sqrt{2} {\mu}_{2}

For

e \to f

, an angle of refraction is lesser than the angle of incidence. So,

{\mu}_{2} > {\mu}_{1}

and then

{\mu}_{2} > {\mu}_{3}

.
For

e \to g, {\mu}_{1} = {\mu}_{2}

.
For

e \to h, {\mu}_{2} < {\mu}_{1} < \sqrt{2}{\mu}_{2}

and

{\mu}_{2} > {\mu}_{3}

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